1216走迷宫

1216走迷宫

⭐️难度：简单
🌟考点：bfs
📖

📚

import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int m = sc.nextInt();
        int [][] a = new int [n + 1][m + 1];

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                a[i][j] = sc.nextInt();
            }
        }
        int x1 = sc.nextInt();
        int y1 = sc.nextInt();
        int x2 = sc.nextInt();
        int y2 = sc.nextInt();

        boolean [][] v = new boolean[n+2][m+2];// 是否被访问
        int [][] d = new int[n+1][m+1]; // 起点到各个点的距离
        for (int i = 1; i <= n; i++) {
            Arrays.fill(d[i],(int)1e9); // 先全部设为无限大，不可到达
        }
        Queue<int[]> q = new LinkedList<>(); // 队列记录即将遍历的结点

        // bfs
        d[x1][y1] = 0;
        q.add(new int[]{x1,y1});
        int[] dx = {0,0,1,-1}; // 分别对应：右左上下 四个操作
        int[] dy = {1,-1,0,0};
        while(!q.isEmpty()){
            int []xy = q.poll();
            int x = xy[0];
            int y = xy[1];
            if(v[x][y]) continue; // 已经被访问过了
            v[x][y] = true;
            for(int i = 0;i < 4;i++){
                int nx = x + dx[i];
                int ny = y + dy[i];
                if(nx > n || nx < 1 || ny > m || ny < 1 || a[nx][ny] == 0 || v[nx][ny]) continue; // 该点不合法：超出范围或不可访问或已被访问
                q.add(new int[]{nx,ny});
                d[nx][ny] = d[x][y] + 1; // 在原来的步数上+1，就走到了当前这个结点
            }
        }
        if(!v[x2][y2]){ // 没有被访问，不可到达
            System.out.println("-1");
        }else{
            System.out.println(d[x2][y2]);
        }
    }
}

细节：Boolean和boolean不一样，Boolean不会初始化为false，会报空指针异常。